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For known $p$, use nearest-integer stopping for amplitude amplification. SetWriting $x=\pi /(4\theta )$, the difference between $r+1/2$ and $x$ has absolute value at most $1/2$. Thus $|(2r+1)\theta -\pi /2|\le \theta$. The Born rule and part 3 giveCount the initial $A$ once and the $A^{†}$ and $A$ occurring inside each of the $r$ amplification iterates. The number of applications of either preparation unitary is thereforeThe counts separately are $r+1$ applications of $A$ and $r$ applications of $A^{†}$, in addition to $r$ applications each of $V$ and $V_0$. This is a quadratic improvement over the $1/p$ mean number of unamplified trials. At an exact rounding tie either nearest iterate meets the probability bound. Knowledge of $p$ is used in selecting the stopping time; repeatedly applying $Q$ without a suitable stopping rule can rotate the state past the good axis and reduce success again. The fact that answers lie in NP supplies verification, but by itself gives no claim that $1/\sqrt{p}$ is polynomial in input length.

Compute the honest one-qubit density operators in the computational basis. Writing $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|u⟩=(5|0⟩+|1⟩)/\sqrt{26}$, the two quantum state ensembles giveThe equality includes the off-diagonal coherence; comparing only the basis outcome probabilities would not suffice.

The protocol is perfectly hiding against Bob when Alice is honest. Independence of the honest preparations makes his full received density operator ${\rho }_0^{\otimes N}={\rho }_1^{\otimes N}$. For any POVM $\{E_z\}$, including collective measurements and measurements using ancillas, the Born rule gives the same outcome probabilities $\mathrm{Tr}(E_z{\rho }_0^{\otimes N})=\mathrm{Tr}(E_z{\rho }_1^{\otimes N})$. Thus no measurement supplies information about the committed bit.

It is not binding against Alice. Here is an explicit ensemble-steering attack on quantum bit commitment. For each position, Alice prepares the normalized two-qubit pure stateretains $A$ and sends $B$. Its partial trace on $A$ is the honest common density operator. She sends the $B$ halves of $N$ independent copies while retaining all the $A$ halves, and chooses her bit only at unveiling.

To open one, she makes a projective measurement of each retained qubit in the computational basis. Outcome zero has probability $13/20$ and prepares Bob's state $|u⟩$; outcome one has probability $7/20$ and prepares $|1⟩$. She announces that list. It has exactly the honest bit-one distribution, and every declared-state test passes.

To open zero, she instead measures each retained qubit using a three-outcome POVM with effects $E_j=|v_j⟩⟨v_j|$, whereAll effects are positive. Their diagonal sums are $4/5+1/5=1$ and $4/70+16/70+5/7=1$, and the off-diagonal sum is zero, so $E_0+E_{+}+E_1=I$. This establishes that the proposed POVM is physically valid.

To verify the conditional states directly, the coefficient matrix of $|\Omega ⟩$, with Bob's index as the row and Alice's as the column, isA real rank-one effect $|v_j⟩⟨v_j|$ on Alice's half gives the unnormalized Bob state vector $C{v}_j$. For the three effects above,Their squared norms are therefore $1/2,1/4,1/4$, exactly the honest bit-zero probabilities, and Bob's conditional pure states are the corresponding declared states. Independent measurements on the retained halves produce the honest distribution of the entire list, so even checks on the declared frequencies do not defeat the attack. Every declared-state projector test again succeeds.

Alice can consequently unveil either bit with acceptance probability one, without having chosen it when sending the qubits. This is the Hughston–Jozsa–Wootters theorem mechanism implemented by an explicit measurement, rather than merely an appeal to the general impossibility of perfectly hiding and binding quantum bit commitment.

Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.

Use the symmetric lossless beam splitter convention with transmission amplitude $t=1/\sqrt{2}$ and reflection amplitude $r=i/\sqrt{2}$. The two mirrors contribute the same fixed quantum phase, which can be suppressed. At the rightward port $D_1$, the upper-arm route has one reflection at BS1 and one transmission at BS2, while the lower-arm route has one transmission at BS1 and one reflection at BS2. Their probability amplitudes therefore add asBy the Born rule,For the upward port, the two amplitudes are $r^2{e}^{i\theta }$ and $t^2$, giving $a_0=(1-e^{i\theta })/2$ and $P(D_0)={\sin }^2(\theta /2)$. They sum to one, as required for an ideal lossless Mach-Zehnder interferometer. Thus $D_1$ is bright at zero relative quantum phase and dark at quantum phase $\pi$. A fixed unequal reflection quantum phase or a swapped port labeling would shift the fringe; the stated convention fixes the port and quantum phase calibration used below.

Without the upper-arm rotation, the photon polarization still labels the path. Before BS2 the quantum state isAfter the balanced beam splitter, the polarization-valued output amplitudes areThe detectors do not resolve photon polarization. Since $⟨H|V⟩=0$, the cross terms vanish in the Born rule:These probabilities are independent of the tunable quantum phase and the input photon polarization, so there is no usable wave interference fringe. Equivalently, taking the partial trace over photon polarization removes the quantum coherence between the two path quantum states. The which-path information is retained in orthogonal photon polarizations even without anybody measuring it. Rotating the two occupied photon polarization quantum states to the same quantum state removes that distinction and allows the cross term in part (ii).

Perfect identification on every trial is impossible, because the family contains a nonorthogonal pair. Any physical identification device induces a POVM whose effect $E_i$ corresponds to its declaration of label $i$. Certainty of correct identification would require $⟨{\psi }_i|E_i|{\psi }_i⟩=1$ and $⟨{\psi }_j|E_i|{\psi }_j⟩=0$ for every $j\ne i$.

Since $0\le E_i\le I$, the first identity gives $(I-E_i)|{\psi }_i⟩=0$, while positivity and the second identity give $E_i|{\psi }_j⟩=0$. For example these conclusions follow by writing the zero expectations as squared norms of $(I-E_i{)}^{1/2}{\psi }_i$ and $E_i^{1/2}{\psi }_j$. ThusPerfect identification would force the entire family to be orthogonal, contrary to the hypothesis. This argument permits ancillas, general measurements and disturbance of the input; it does not assume that the device must preserve the state.

Unambiguous identification with a nonzero success probability for every input is possible. Normalize the given kets and let $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ be their Gram matrix. Their linear independence makes $G$ positive definite. Define reciprocal vectors within their span byThen $⟨{\psi }_k|{\chi }_i⟩={\sum }_j{G}_{kj}(G^{-1}{)}_{ji}={\delta }_{ki}$, and hence also $⟨{\chi }_i|{\psi }_k⟩={\delta }_{ik}$. Put $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$. It is positive and has finite operator norm. Choosing $ϵ=1/(2\Vert R\Vert )>0$, setAll these effects are positive, and $E_{?}\ge I/2$, so they sum to the identity and define a POVM. The Born rule givesThus every label is correct whenever it occurs and every input has the strictly positive success probability $p_k=ϵ$. This is the reciprocal-state construction of unambiguous discrimination.

For an explicit success constant, collect the input kets into a matrix $\Psi$, so ${\Psi }^{†}\Psi =G$. Its polar factorization is $\Psi =U{G}^{1/2}$ with $U^{†}U=I_N$. The reciprocal columns are $\Psi G^{-1}=U{G}^{-1/2}$, and therefore $R=U{G}^{-1}{U}^{†}$. Its largest eigenvalue is $1/{\lambda }_{\min }(G)$, so one can take $ϵ={\lambda }_{\min }(G)/2$. Since the normalized Gram matrix has trace $N$, this is at most $1/2$.

To realize the device, use measurement operators $M_i=\sqrt{E_i}$ and $M_{?}=\sqrt{E_{?}}$. The map $|v⟩\mapsto {\sum }_i{M}_i|v⟩|i⟩+M_{?}|v⟩|?⟩$ preserves inner products because the effects sum to $I$. Extend this linear isometry to a unitary on the input and an initially blank ancilla, then projectively measure the ancilla's orthogonal outcome labels. Its outcome probabilities are precisely those above. This gives a physical unambiguous quantum state discrimination device rather than only a formal set of effects.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

With the negative exponential convention of the specified quantum Fourier transform, the amplitude of computational outcome $k$ isIf $k=m$, every summand is one. Otherwise this finite geometric series has ratio $q\ne 1$ with $q^N=1$, so it equals $(1-q^N)/(1-q)=0$. ThusThe Born rule gives outcome $m$ with probability one when performing a quantum measurement in the computational basis. Recover the promised phase as ${\varphi }_m=2\pi m/N$ modulo $2\pi$. Using the opposite Fourier sign would instead return the label $-m\mathrm{mod}N$, so the sign convention matters.

For the Kraus operators of the measurement in quantum mechanics, the Born rule givesThe density matrix $A_i\rho A_i^{†}$ is the unnormalized state of that outcome. It is positive, its trace is $p_i$, and completeness gives ${\sum }_i{p}_i=\mathrm{Tr}\rho =1$. There is no conditional state to assign to an event with $p_i=0$.

For a projective measurement, each $P_i$ is an orthogonal projection, satisfying $P_i=P_i^{†}=P_i^2$. Completeness becomes ${\sum }_i{P}_i=I$. Fix $v$ in the range of $P_j$. ThenSince $P_{j}v=v$, the $j$th summand already equals $\Vert v{\Vert }^2$. All other summands are nonnegative, so $P_{i}v=0$ for $i\ne j$. Consequently $P_i{P}_j=0$, and taking adjoints also gives $P_j{P}_i=0$. This proves the stronger completeness forces orthogonality of measurement projections:

For the unitary dilation of a measurement instrument, take a quantum ancilla with orthonormal states $|0⟩,|1⟩,\dots ,|n⟩$, initially in $|0⟩$. DefineThe completeness equation impliesso $V$ is an isometry. Its action on the input subspace can be extended to a unitary operator $U$ on the enlarged system:Choose orthonormal bases in the two complementary subspaces and map one to the other. Their dimensions agree; the extra unused quantum ancilla level also ensures this extension is possible for an infinite-dimensional system. After the coupling, the joint density matrix is

Now perform a projective measurement on the larger system, withThese are pairwise orthogonal and sum to the full identity. The unused $|0⟩$ level has zero weight in $\Omega$, so merging it into outcome one adds no probability. For every $i$,Its trace is $p_i$, and taking the partial trace over the quantum ancilla after normalization gives exactly ${\rho }_i$. Hence unitary coupling to an quantum ancilla, a projective measurement, and discarding the quantum ancilla reproduce both the probabilities and the conditional states of the general measurement. This is the instrument version of a Stinespring dilation, rather than a construction reproducing only the outcome probabilities.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

The orthogonal projection onto the $+1$ eigenspace of the Pauli matrix ${\sigma }_z$ is $P_{+}=(I+{\sigma }_z)/2$. By the Born rule and the Bloch vector representation,Since $s_z=1/5$,The given vector has length $19/30<1$, so it lies inside the Bloch ball and indeed defines a valid mixed qubit state. For the physical spin operator $S_z=(\hslash /2){\sigma }_z$, the corresponding outcome is $+\hslash /2$ with the same probability.

For one fixed POVM repeated on independently prepared copies, the Born rule gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit density operator has three independent real parameters,The probability map is affine in $\mathbf{r}$. It can be injective on the full Bloch ball only if $m-1\ge 3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\ge 4$ for an informationally complete POVM.

To attain this bound, take the four directionsThey are the vertices of a regular tetrahedron. The tetrahedral qubit POVM has effects $E_i=(I+{\mathbf{n}}_i\cdot \mathbf{\sigma })/4$, each positive and rank one, and ${\sum }_i{E}_i=I$. The observed probabilities areSince ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I_3/3$, they recover the state byThis is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is informationally complete three-observable qubit tomography, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Set $L=2^N$. Applying the inverse quantum Fourier transform to the register in part (a) gives probability amplitude at label $m$ equal toChoose $m$ nearest to $\eta =L{\varphi }_j/(2\pi )$, with $m$ reduced modulo $L$ at the phase boundary. Then the difference can be represented with $|\delta |\le \pi /L$. For $\delta \ne 0$, sum the finite geometric series to obtainLet $z=L|\delta |/2\le \pi /2$. On $[0,\pi /2]$, concavity of sine gives $\sin z\ge 2z/\pi$, the line through its endpoint values. Also $|\sin (\delta /2)|\le |\delta |/2$. Thereforeand the Born rule proves the nearest-integer success bound for quantum phase estimation:If $\delta =0$, every summand is one and the probability is exactly one. If the scaled phase is halfway between two integers, either nearest label separately satisfies the displayed bound. Interpreting the measured integer as $m/L$ gives the closest available $N$-bit phase estimate modulo one.

Here is a tripartite Schmidt projection entanglement concentration protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local qubits of each party giveswhere $|x|$ counts ones. Alice performs a projective measurement of that count, without learning the individual string. The orthogonal projector for outcome $k$ is ${\Pi }_k={\sum }_{|x|=k}|x⟩⟨x|$. By the Born rule, its probability and conditional pure state areAlice sends $k$ by classical communication; all three parties use local unitary operations to relabel the same ordered type class as $0,\dots ,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual qubits instead would have destroyed precisely this coherence.

To obtain an integer number $f$ of target GHZ states, choose $m=2^f\le d_k$. The following exact LOCC step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq \{0,\dots ,d_k-1\}$, Alice uses the Kraus operatorEvery label occurs in $\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)$ such subsets, proving ${\sum }_S{K}_S^{†}{K}_S=I$ on the occupied subspace. Thus this is a complete measurement in quantum mechanics. Each outcome has probability $m/[d_k(\genfrac{}{}{0ex}{}{d_k-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d_k}{m}\right)$, and leaves the normalized pure state $m^{-1/2}{\sum }_{r\in S}|r,r,r⟩$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into exactly $f$ independent copies of the target GHZ state. Local extensions of these operations handle unused basis vectors and do not affect the protocol.

For example, taking $f(k)=\lfloor {\log }_2{d}_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a binomial distribution with parameters $(n,q)$, the weak law of large numbers gives $K/n\to q$ in probability. Stirling's formula, uniformly when $k/n$ stays away from zero and one, givesBy continuity of the binary entropy, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.

A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, takeOn $|K-nq|\le n^{2/3}$, the bounded derivative of the binary entropy in a neighborhood of the fixed $q\in (0,1)$ and Stirling's formula give ${\log }_2{d}_K\ge n{h}_2(q)-C_q{n}^{2/3}-O(\log n)\ge f_n$. Chebyshev's inequality bounds the complementary probability by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset measurement in quantum mechanics with $m=2^{f_n}$. ThusThis matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.

Hermiticity of the Hamiltonian operator gives $⟨{\psi }_m,H{\psi }_n⟩=⟨H{\psi }_m,{\psi }_n⟩$, soThe distinct eigenvalues force orthogonality for $m\ne n$, and the given normalization gives one for $m=n$. Therefore $⟨{\psi }_m,{\psi }_n⟩={\delta }_{mn}$.

Under the time-independent Schrödinger equation, each energy component acquires its own phase:The sum converges in the state Hilbert space because the squared coefficients are summable. Orthogonality makes its squared norm ${\sum }_{n\ge 0}{2}^{-(n+1)}=1$ at every time.

The Born rule for energy measurement givesBoth depend only on squared coefficient magnitudes, so both are time-independent. The expectation follows either by summing energy times measurement probability or by inserting the orthogonal expansion in the quadratic form of $H$.

Normalization alone does not guarantee a finite mean for an arbitrary spectrum. The displayed energy series is finite when $\sum |E_n|2^{-(n+1)}<\infty$. Since the spectrum is bounded below by $E_0$, a divergent positive tail gives an extended expectation $+\infty$, still independent of time; for example an abstract diagonal Hamiltonian with $E_n=4^n$ has this property. No growth assumption on $E_n$ is printed, so finiteness should not be asserted without this qualification.

Let $R=|\psi ⟩⟨\psi |$ and choose an orthonormal basis $\{|e_{\mu }⟩\}$ of ${\mathcal{H}}_2$. The reduced density matrix on $S_1$ is the partial traceHere $I_1$ is the identity on ${\mathcal{H}}_1$, and the bras and kets in the sum contract only the second tensor factor. Equivalently, if $|\psi ⟩={\sum }_{a,\mu }{c}_{a\mu }|f_a⟩\otimes |e_{\mu }⟩$, then $({\rho }_1{)}_{ab}={\sum }_{\mu }{c}_{a\mu }{c}_{b\mu }^{\ast }$. The definition is independent of the chosen basis: it is characterized by $\mathrm{Tr}(M{\rho }_1)=\mathrm{Tr}[(M\otimes I_2)R]$ for every operator $M$ on ${\mathcal{H}}_1$.